Starting from Maxwell's equations, deduce Faraday's law of induction

$$\frac{d \Phi}{d t}=-\varepsilon$$

for a moving circuit $C$, where $\Phi$ is the flux of $\mathbf{B}$ through the circuit and where the EMF $\varepsilon$ is defined to be

$$\varepsilon=\oint_{C}(\mathbf{E}+\mathbf{v} \times \mathbf{B}) \cdot d \mathbf{r}$$

with $\mathbf{v}(\mathbf{r})$ denoting the velocity of a point $\mathbf{r}$ of $C$.

[Hint: consider the closed surface consisting of the surface $S(t)$ bounded by $C(t)$, the surface $S(t+\delta t)$ bounded by $C(t+\delta t)$ and the surface $S^{\prime}$ stretching from $C(t)$ to $C(t+\delta t)$. Show that the flux of $\mathbf{B}$ through $S^{\prime}$ is $-\oint_{C} \mathbf{B} \cdot(\mathbf{v} \times d \mathbf{r}) \delta t$.]